Inverse Square Law

I have a photo booth coming up. First one. I have decided to keep things simple and use one shoot through umbrella (centred), flash on manual and set up the initial exposure via light light meter. Flash will be a 580 II.

Since by that time of the celebration half the people will be half in the bag so trying to control them will be next to impossible. I plan to put a piece of tape across the floor and tell them just not to cross it.

What would be a comfortable distance to subject based on the flashes power so as they are goofing around I have about a 4 foot range where everyone is evenly lit.

I found a couple of diagrams in the net but one is in metric and one in feet. The one in feet is 4, 8 and 12 which seems to contradict the metric range. I finally gave up and thought I'd throw this question out here. I looked for hours and could not find a chart in feet that was more detailed than 4, 8 and 12.

By metric I thought a 2 meters or 6 feet.

Is there a question in there somewhere? There is no question mark (?).
Looks like you have it all figured out!

An easy way to remember power settings for a flash and modifier is to set the flash and modifier on a lightstand 11 feet from the subject. Set the Flash to 1/4 power, camera to ISO 400 and lens to F8 make an exposure that includes a crumpled white paper towel or similar white reference. Look at the back of your camera to see if there are any blinkies on the LCD in the white. If there are close down a stop and try again. If no blinkies open up 1/3 stop and try again. Repeat until you first see blinkies then close down 1/3 stop. That is the optimum exposure for the flash and modifier at 1/4 power at 11 Ft.

The reason the stand is set at 11 ft? Because now you can move the stand to 8 ft and know that you have to stop down the lens one stop or reduce power to 1/8th. Obviously it works the other way as well. And ISO and F stop on camera are also options.

As for your question, you need to have the umbrella back far enough so there will not be a dramatic exposure change for a couple of feet so I'd say at least 10 feet. That may be pushing the 580EX a bit through an umbrella since you will also want to be stopped down a bit for some DoF.

Kechar wrote in post #12586469
Is there a question in there somewhere? There is no question mark (?).
Looks like you have it all figured out!

Yes looks like I did not ask any questions. I thought about 6 feet to subject but I was not sure. I should have had one there.

dmward wrote in post #12586477
An easy way to remember power settings for a flash and modifier is to set the flash and modifier on a lightstand 11 feet from the subject. Set the Flash to 1/4 power, camera to ISO 400 and lens to F8 make an exposure that includes a crumpled white paper towel or similar white reference. Look at the back of your camera to see if there are any blinkies on the LCD in the white. If there are close down a stop and try again. If no blinkies open up 1/3 stop and try again. Repeat until you first see blinkies then close down 1/3 stop. That is the optimum exposure for the flash and modifier at 1/4 power at 11 Ft.

The reason the stand is set at 11 ft? Because now you can move the stand to 8 ft and know that you have to stop down the lens one stop or reduce power to 1/8th. Obviously it works the other way as well. And ISO and F stop on camera are also options.

As for your question, you need to have the umbrella back far enough so there will not be a dramatic exposure change for a couple of feet so I'd say at least 10 feet. That may be pushing the 580EX a bit through an umbrella since you will also want to be stopped down a bit for some DoF.

OK I'll try at 10 feet as I'm setting up. 8 should work as well. I was planning on F8. Thanks for giving me the info.

The drawing is wrong.
The correct distance for f/4 would be 2.828427125 meters (2 * square root of 2), not 3 meters as the drawing suggests, assuming the other distances are correct.

The above chart/drawing could use units of nanometers, feet, miles, furlongs or cubits. The unit of measurement doesn't matter. The theory is what you need to understand.

The theory is simple: Every time you double the distance, you quadruple the area illuminated by a given number of photons. Since a given area receives one-fourth as much light, you need 2 stops more aperture (or ISO, or shutter speed if the light source is continuous) for equivalent exposure.

Now here's the cool part about it: Since aperture also relates to exposure in an exponential way (f/4 is 1/4 the exposure of f/2), the distance multiplied by the aperture for a given light source will always be the same. This is the guide number often used to measure flash power.
In the scenario above, the light has a guide number of 11 meters.
1 * 11 = 11
2 * 5.6 = 11
2.8 * 4 = 11
4 * 2.8 = 11.

So here's how you use it in the field. Suppose your light is 10 feet from subject and you get a reading of f/5.6
Distance * aperture = 10 * 5.6 = 56 feet.
Now you move the light in so it's 6 feet from subject.
56/6 = 9.3 so an aperture setting of f/9.2 will pretty much nail it, without having to meter again.

Ain't that cool?

Kechar wrote in post #12586469
Is there a question in there somewhere? There is no question mark (?).
Looks like you have it all figured out!

Actually what I was looking for is the percentage of drop off as well which I should haev aked for as well. I found this and it made it more clear as he uses a 1 to 10 foot example and then to 16. This is a pretty video. I like Mark Wallace's explanations.

http://fstoppers.com …e-easy-inverse-square-law

Recalling my physics course a few weeks ago (:rolleyes, the inverse square law only works as advertised with a point source of light. When using a large light source (relative to the light-to-lit-objects distance), the falloff does not exactly follow the inverse square law.

Skip, you are correct, as usual.

Since, in reality, there's no such thing as a point source, you could say the inverse square law never really works.

With a hotshoe flash at 20 feet, it works for our purposes.

With a 60 inch softbox at 2 feet, it pretty much falls apart.

So we have the inverse square law for rough estimates, and light meters for those times when we need precision.

That makes sense. Interesting. All relative to the set up like everything else.

too much math for me. I'd rather chimp and watch my histogram:p

Yep - pass the bananas over here as well. With this photo booth I just want to be sure that in case people goof around and get 2 deep the front aren't blow out and the back dark. Not that it will be that bad. I hate entering something and being unprepared for it.

Besides I don't think it is as much as math as much as knowing about and using it to your advantage or better not getting into trouble with it. I may get a response to to math thing. I can't see myself do the math part as I would just use my light meter.

Sorry - gotta get back to the Bruin's. Been a fan for over 45 years.

Curtis N wrote in post #12588154
With a hotshoe flash at 20 feet, it works for our purposes.

With a 60 inch softbox at 2 feet, it pretty much falls apart.

I fully agree with those examples.

In the case of flash (virtual 'point source' of light) the 'math' is simple division guys, 5th grade arithmetic (after all, decimal digits are involved, not whole numbers!!!) sheesh

GN = distance * f/stop, or
GN / distance = f/stop

The difficulty with today's flashes is that as the head zooms to different FL, the GN changes. In the case of 580EXII, the GN190 only applies to 105mm FL coverage, and reduces to GN130 at 50mm ('normal' lens for 135 format). So GN arithmetic is not as simplistic as it was before zoom head flash units were invented!

As pointed out, softboxes do not follow 'Inverse Square' at close distances (where distance to softbox is about 2-3x the largest dimension of the softbox)...it is Inverse Linear. Best to simply use a flash meter, rather than to use Guide Number arithmetic!

A real easy crutch is simply to think of f/numbers as distances...if you have f/4 at 11', you have f/11 at 4'. Or if you have f/4 at 11', you have f/5.6 at 8'...see the relationships in these simple examples here?! OK, now to make it just a bit more difficult (making the distances less convenient, but still keeping the concept simple)....if you have f/4 at 17' you have f/2 at 34' or f/8 at 8.5' Still only using 5th grade arithmetic here.

Using f/stops like distances, if my flash is positioned at 4' from the subject and measures f/6.9, I could decrease its power by -1EV by moving the flash from 4' to 5.6', or decrease the flash power by -2EV by moving it from 4' to 8'. Do the numbers 4, 5.6, and 8 look familar?...they should, f/4, f/5.6, f/8. Not even 5th grade arithmetic here!

Thanks for the detailed explanation Wilt.